AP Calculus AB and BC

Limit of (e^x - e^-x) / x at 0 Is 2

The limit of e to the x minus e to the minus x, all over x, as x approaches 0 is 2. Splitting it gives two copies of the standard limit, one worth 1 and the other worth 1 again because the two minus signs multiply to a plus.

lim⁡x→0ex−e−xx=2\lim_{x \to 0} \frac{e^{x} - e^{-x}}{x} = 2

Settled by splitting into two standard exponential limits.

Splitting into two known limits

Add and subtract 11 in the numerator so that each exponential gets its own copy of the standard limit.

ex−e−xx=(ex−1)−(e−x−1)x=ex−1x−e−x−1x\frac{e^{x} - e^{-x}}{x} = \frac{\left(e^{x} - 1\right) - \left(e^{-x} - 1\right)}{x} = \frac{e^{x}-1}{x} - \frac{e^{-x}-1}{x}

The first term tends to 11. The second is the same limit with −x-x in place of xx, which contributes −1-1, and subtracting it adds 11.

lim⁡x→0ex−e−xx=1−(−1)=2\lim_{x \to 0}\frac{e^{x} - e^{-x}}{x} = 1 - (-1) = 2

Or read it as a derivative

The expression is the derivative of e^x minus e^-x at 0. That derivative is e^x plus e^-x, which is 2 at the origin. Same answer, one line.

Why the sign works out to addition

Two minus signs are in play and they multiply. Differentiating e−xe^{-x} produces a −1-1 from the chain rule, and the term is already being subtracted, so the two negatives combine into a positive contribution.

Compare the version with a plus sign in the middle: ex+e−x−2x\frac{e^{x} + e^{-x} - 2}{x} tends to 00, because there the contributions cancel instead of reinforcing.

The mistakes students make

  • Answering 00 by assuming the two exponentials cancel. At small xx they are nearly equal, but their DIFFERENCE is about 2x2x, which is exactly the size of the denominator.
  • Losing the chain rule sign on e−xe^{-x} and answering 00.
  • Cancelling the xx into the exponent. The xx downstairs is a factor, not something that can be moved into a power.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of (e^x - e^-x) / x as x approaches 0?

It is 22.

Why is the answer not 0?

The two exponentials are close but not equal: their difference is about 2x2x near the origin, matching the denominator exactly.

What is the L'Hopital route?

One pass gives ex+e−x1\frac{e^{x} + e^{-x}}{1}, which is 1+1=21 + 1 = 2 at the origin.